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Help with a derivative

  1. Jun 30, 2004 #1
    I am having trouble with part of a problem

    what is d/dx (sec 5x)

    is it ((sec 5x)(tan 5x)(d/dx 5x)) ?

    Or is 5 a constant multiple ?

    either way what next ???

    What would (sec5x)(tan 5x)(5) be?

    Thanks
     
  2. jcsd
  3. Jun 30, 2004 #2
    Remember that [itex]d/dx~f(g(x)) = f'(g(x))~g'(x)[/itex].
    Simplify and find out.
     
  4. Jun 30, 2004 #3
    How about

    d/dx (sec 5x)

    = (sec 5x)(tan 5x)(d/dx 5x)
    = (sec 5x)(tan 5x)(5)
    =(sec 5x)(5tan 5x)

    I am using the formula d/dx sec u = (sec u)(tan u)(d/dx u)

    Am I on the right track?
     
  5. Jun 30, 2004 #4
    maybe

    x(sec 5)(5tan 5)
     
  6. Jun 30, 2004 #5
    Write out the sec and tan as functions of sin and cos if you want to simplify properly.
     
  7. Jul 1, 2004 #6
    software

    Does anyone know a program available I can use to check my answers to problems like these...quotient rule, product rule, and chain rule.

    Thanks
     
  8. Jul 2, 2004 #7
    Some symbolic algebra program should help...how about Wolfram Research's Mathematica? (www.wolfram.com)

    I am not yet aware of anything in the GNU which does the same job....


    Cheers
    Vivek
     
  9. Jul 2, 2004 #8
    A TI-89 graphing calculator will also solve many of these problems. Invaluable for checking your work.
     
  10. Jul 13, 2004 #9
    Hello everyone,
    I'm having some difficulties with a problem.
    It includes an equation and says find the time when it is a maximum.
    The equation is f(x)=75/(1+74e^(-0.6t))
    To find the maximum we'll have to set the derivative of f(x)=0, but the problem I have is differentiating this expression. Applying the quotient rule, (75 is a constant) I get stuck at differentiating 1 + 74e^(-0.6t). It is simple using logarithms to differentiate it. For example y= 74e^(-0.6t) (d/dx 1= 0)
    ln y= 74*-0.6t ln e y'/y=-44.4t *1 y'= -44.4t* 74e^(0.6t)
    Now, I am not sure if I can use this technique when finding the derivative of the whole expression f(x)=75/(1+74e^(-0.6t)) and then setting it equal to 0?
    Thank you
    K.K
     
  11. Jul 14, 2004 #10
    Yeah, there ain't no problem with using the rule of quotients with the deriv. you just calculated. The quotient rule is independant of how you calculate each derivative. By the way, there ain't no need to use logarithmic derivative since the derivative of exp() is soooo easy.
    [tex] d\frac{ 1 + 74e^{-0.6t}} {dt} = 74* (-0.6) e^{-0.6t}[/tex]

    Anyway, be aware that (f ' =0) is only a sufficient codition for maximum (or minimum, for that matter), so you should check the second derivative as well.
     
    Last edited: Jul 14, 2004
  12. Jul 14, 2004 #11
    By the way, the total deriv. is:
    [tex]\frac{3330 e^{-0.6t}}{(1 + 74 e^{-0.6 t})^2}[/tex]
     
  13. Jul 14, 2004 #12
    Fortunately I was right, anyway thanks for your help. I found the derivative exactly as you did. Now I should find the maximum. So i should find f(x)' = 0 and f(x)" if it is <0 (max) or > 0 (min). The problem I have now is that I can't find a value for f(x)' (a value for t) so that it is equal to 0? :cry:
     
  14. Jul 14, 2004 #13

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The function you give, f(x)=75/(1+74e^(-0.6t)), is steadily increasing. It has no maximum. If your problem asks for the maximum on a closed interval, then it is the value at the right end point.,
     
  15. Jul 14, 2004 #14
    More generally if f = sec u and u = 5x,

    [tex]
    \frac{df}{du} = \sec u \tan u
    [/tex]

    and

    [tex]
    \frac{du}{dx} = 5
    [/tex]

    The chain rule says that

    [tex]
    \frac{df}{dx} = \frac{df}{du} \frac{du}{dx}
    [/tex]

    so that [tex]\frac{df}{dx}[/tex] is easily found out as [tex]5 \sec 5x \tan 5x [/tex].

    Whether you use product rule or take logs first, you should end up with the same answer. Which trick to use when is an idea you get only after solving too many problems...sometimes log is better and it avoids too much algebra.

    Remember the quotient rule as

    [tex]
    \frac{d}{dx}[\frac{f(x)}{g(x)}] = \frac{g(x) \frac{df(x)}{dx} - f(x) \frac{dg(x)}{dx}}{[g(x)]^{2}}
    [/tex]

    that is first write the square of the denominator in the denominator on the right hand side. Then write the expression in the numerator as: (denominator of (f/g) * derivative of numerator (df/dx)) - (numerator of (f/g) * derivative of denominator (dg/dx)).

    For your problem (f/g) or (f(x)/g(x)) represents the function to be maximized/minimized, so using the above form of the quotient rule, you can directly set d(f/g)/dx equal to zero, as

    [tex]
    g(x) \frac{df(x)}{dx} - f(x) \frac{dg(x)}{dx}} = 0
    [/tex]

    Fill in the rest ...
     
    Last edited: Jul 14, 2004
  16. Jul 20, 2004 #15
    L'Hospital Rule

    Using L'Hospital's Rule I should find the limit of the function
    Lim x->inf. x[ln(x+5) -ln x] . First it should be differentiated.
    I wrote the function like this : Lim x->inf. x ln(x+5) - x ln x or
    Lim x->inf. [ln(x+5)]/(1/x) - ln x/(1/x). I differentiated the function and the 1st derivative is -x^2/(x+5) +x . Even trying with second derivative I can't find the correct answer. Maybe there's sth wrong?
     
  17. Jul 20, 2004 #16
    No, it's perfectly correct. -x^2/(x + 5) + x = (-x^2 + x(x + 5)) / (x + 5) = (-x^2 + x^2 + 5x) / (x + 5) = 5x/(x + 5), apply l'Hopitals rule again to find the limit.
     
  18. Jul 20, 2004 #17
    :approve:
    I am sorry for inconvenience. I forgot that ln(x+5) - ln(x) = ln (x+5)/x. The exercise was really easy.
     
  19. Jul 26, 2004 #18
    Optimization

    "A conical cup is made from a circular piece of paper of radius R by cutting out a sector and joining its edges. Find the maximum capacity of such cup"

    I just know that V(cone)=1/3 Pi R^2 H and that the lateral Area is A= Pi R Sqr(R^2+H^2)

    I can't go any further.
    Thanks for the help
     
  20. Jul 26, 2004 #19
    I suggest you start a new thread next time. I'm not sure how exactly this conical cup is being formed. You say "cutting out a sector and joining its edges". How big a sector are you talking about? Does 'its' refer to the sector or the circular piece of paper without the sector? What is the shape of this sector (I'm guessing it looks like a piece of pie)?
     
  21. Jul 26, 2004 #20
    Yes, do start a new thread.

    e(ho0n3, http://mathworld.wolfram.com/CircularSector.html

    Electro, you know that the original radius of the circle is going to be the slant height and that that circumference of the base of the cone is going to be the same as the original cirumference minus the part that was taken out with the sector. How can you use this?

    cookiemonster
     
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